[HW] OJ记录20题之一
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注意
int main(){ int n; cin>>n; while (n--) { char ch[100]; cin>>ch; //此处用gets(ch)或者cin.getline(ch,100)出现问题 //原因在于输入n后残留了一个回车键，可用getchar()获取。 cout<<fun(ch)<<endl; } return 0;}1 删除字符串中出现次数最少的字符
特殊案例（输入：aaa，输出：NULL）
#include <iostream>#include <cstring>using namespace std;void fun(const char *src, char *dest);int main(){ char ch[21]; while (cin>>ch) { char dest[21] = {0}; fun(ch, dest); cout<<dest<<endl; } return 0;}void fun(const char *src, char *dest){ int c[26] = {0}; int least = 20; for (int i = 0; i < strlen(src); i++) { c[src[i]-'a']++; } for (int i = 0; i < 26; i++) { if (c[i] != 0) { least = min(least, c[i]); } } while ('\0' != *src) { if (c[*src-'a'] != least) { *dest++ = *src; } src++; } *dest = '\0';}2 输入n个整数，输出其中最小的k个
#include <iostream>#include <cstring>using namespace std;bool GetMinK(unsigned int uiInputNum, int * pInputArray, unsigned int uiK, int * pOutputArray);int main(){ int uiInputNum, uiK; while (cin>>uiInputNum>>uiK) { int *pInputArray = new int[uiInputNum]; for (int i = 0; i < uiInputNum; i++) { cin>>pInputArray[i]; } int *pOutputArray = new int[uiK]; GetMinK(uiInputNum, pInputArray, uiK, pOutputArray); for (int i = 0; i < uiK; i++) { if (i != uiK - 1) { cout<<pOutputArray[i]<<" "; } else { cout<<pOutputArray[i]; } } cout<<endl; delete [] pInputArray; delete [] pOutputArray; } return 0;}bool GetMinK(unsigned int uiInputNum, int * pInputArray, unsigned int uiK, int * pOutputArray){ for (int i = 0; i < uiK; i++) { int k = i; for (int j = i + 1; j < uiInputNum; j++) { if (pInputArray[j] < pInputArray[k]) { k = j; } } if (k != i) { int temp = pInputArray[k]; pInputArray[k] = pInputArray[i]; pInputArray[i] = temp; } } for (int i = 0; i < uiK; i++) { pOutputArray[i] = pInputArray[i]; } return true;}3 合法IP
#include <iostream>#include <cstdio>#include <cstring>using namespace std;bool fun(char *str);int main(){ char str[16]; while (cin.getline(str, 16) && strlen(str) != 0) { if (fun(str)) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } } return 0;}bool fun(char *str){ int b[3] = {-1}; int n = 0; int bi = 0; for (int i = 0; i < strlen(str); i++) { if (str[i] == '.') { b[bi++] = i; if (n < 0 || n > 255) { return false; } n = 0; } else { n = n * 10 + str[i] - '0'; } } if (b[0] == 0 || b[2] == strlen(str) - 1 || b[0] == -1 || b[1] == -1 || b[2] == -1 || b[0] == b[1] || b[1] == b[2] || b[2] == b[3]) { return false; } return true;}4 句子逆序
#include <iostream>#include <cstdio>#include <cstring>using namespace std;void reverse(char *str, int begin, int end);int main(){ char ch[100]; while (cin.getline(ch, 100) && strlen(ch) != 0) { reverse(ch, 0, strlen(ch) - 1); int begin = 0, end = 0; for (int i = 0; i < strlen(ch); i++) { if (ch[i] != ' ') { end = i; } else { reverse(ch, begin, end); begin = i + 1; end = i + 1; } } reverse(ch, begin, end); cout<<ch<<endl; } return 0;}void reverse(char *str, int begin, int end){ char *left = str + begin; char *right = str + end; while (left < right) { char ch = *left; *left++ = *right; *right-- = ch; }}5 白钱买白鸡问题
#include <iostream>using namespace std;int main(){ int n; cin>>n; for (int i = 0; i <= 100 / 5; i++) { for (int j = 0; j <= 100 / 3; j++) { int k = 100 - i - j; if (5 * i + 3 * j + k / 3.0 == 100) { cout<<i<<" "<<j<<" "<<k<<endl; } } } return 0;}6 名字的漂亮度
每个字符都有一个漂亮度，求名字的最大漂亮度。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;int fun(char *str);int main(){ int n; cin>>n; while (n--) { char ch[100]; cin>>ch; //此处用gets(ch)或者cin.getline(ch,100)出现问题 cout<<fun(ch)<<endl; } return 0;}int fun(char *str){ int c[26] = {0}; for (int i = 0; i < strlen(str); i++) { if (str[i] >= 'a' && str[i] <= 'z') { c[str[i]-'a']++; } else if (str[i] >= 'A' && str[i] <= 'Z') { c[str[i]-'A']++; } } //sort the count of characters for (int i = 0; i < 26 - 1; i++) { int k = i; for (int j = i + 1; j < 26; j++) { if (c[j] < c[k]) { k = j; } } if (k != i) { char temp = c[k]; c[k] = c[i]; c[i] = temp; } } int sum = 0; int n = 1; for (int i = 0; i < 26; i++) { sum += c[i] * n++; } return sum;}7 字符串分割
输入：
2
abc
123456789
输出：
abc00000
12345678
90000000
（注意：空字符串不处理）
#include <iostream>#include <cstdio>#include <cstring>using namespace std;void fun(char *str);int main(){ int n; cin>>n; while (n--) { char ch[101]; cin>>ch; fun(ch); } return 0;}void fun(char *str){ if (strlen(str) == 0) { return; } int len = strlen(str); while (len % 8 != 0) { len++; } char word[9] = {0}; int index = 0; int i = 0; while ('\0' != *str) { while ('\0' != *str && i < 8) { word[i++] = *str++; index++; } if (i == 8) { word[i] = '\0'; cout<<word<<endl; i = 0; } } if (index < len) { while (index < len) { word[i++] = '0'; index++; } word[i++] = '\0'; cout<<word<<endl; }}8 查找输入二进制整数中1的个数（不使用位运算）
#include <iostream>using namespace std;int fun(int n);int fun2(int n);int main(){ int n; while (cin>>n) { cout<<fun(n)<<endl; cout<<fun2(n)<<endl; } return 0;}int fun(int n){ int count = 0; while (n) { n &= n - 1; count++; } return count;}int fun2(int n){ if (n >= 0) { int count = 0; while (n) { count += n % 2; n /= 2; } return count; } else { return 32 - fun2(-n - 1); }}9 等差数列求和
2 5 8 11 ……
输入：2
输出：7
(用公式求和)
#include <iostream>using namespace std;int main(){ int n; while (cin>>n) { cout<<1.5 * n * n + 0.5 * n<<endl; } return 0;}10 字符串匹配
判断是否一个字符串中所有的字符都在第二个字符串中出现
输入：
ab
abc
输出：
true
#include <iostream>#include <cstdio>using namespace std;bool IsAllCharExist(char* pShortString,char* pLongString);int main(){ char pShort[100]; char pLong[100]; while (gets(pShort) && gets(pLong)) { if (IsAllCharExist(pShort, pLong)) { cout<<"true"<<endl; } else { cout<<"false"<<endl; } } return 0;}bool IsAllCharExist(char* pShortString,char* pLongString){ bool s[256] = {false}; while ('\0' != *pShortString) { s[*pShortString++] = true; } while ('\0' != *pLongString) { s[*pLongString++] = false; } for (int i = 0; i < 256; i++) { if (s[i]) { return false; } } return true;}11 字符统计
统计其中的大小写字母，空格和数字，按个数从多到少输出，若个数一样则先输出SCII码比较小的。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;void fun(char *src, char *dest);int main(){ char src[100]; while (gets(src)) { char dest[100]; fun(src, dest); cout<<dest<<endl; } return 0;}void fun(char *src, char *dest){ int c[123] = {0}; for (int i = 0; i < strlen(src); i++) { if (src[i] < 123) c[src[i]]++; } int index = 0; for (int j = 1; j < 53; j++) //53是所有要统计字符的个数 { int max = c[32]; int curIndex = 32; for (int i = 48; i < 123; i++) { if ((i >= 58 && i <= 64) || (i >= 91 && i <= 96)) { continue; } else { if (c[i] > max) { max = c[i]; curIndex = i; } } } if (max == 0) //统计已经完成 { break; } dest[index++] = curIndex; c[curIndex] = 0; } dest[index++] = '\0';}12 进制转化
输入：0xA
输出：10
#include <iostream>#include <cstdio>#include <cstring>using namespace std;void HEX2DEC(char *src, char *dest);int fun(char c);void itoa(int n, char *str);void reverse(char *str);int main(){ char src[10]; while (gets(src)) { char dest[11]; HEX2DEC(src, dest); cout<<dest<<endl; } return 0;}void HEX2DEC(char *src, char *dest){ int sum = 0; for (int i = 2; i < strlen(src); i++) { sum = sum * 16 + fun(src[i]); } itoa(sum, dest);}int fun(char c){ if (c >= '0' && c <= '9') { return c - '0'; } else if (c >= 'A' && c <= 'F') { return c - 'A' + 10; }}void itoa(int n, char *str){ int i = 0; while (n) { str[i++] = n % 10 + '0'; n /= 10; } str[i++] = '\0'; reverse(str);}void reverse(char *str){ char *left = str; char *right = str + strlen(str) - 1; while (left < right) { char c = *left; *left++ = *right; *right-- = c; }}或
#include <cstdio>int main(){ int n; scanf("%x", &n); printf("%d", n); return 0;}13 提取不重复整数
从右边开始往左边读，如果以0开头，则去掉0。
输入：
0000
123355660
输出：
0
65321
#include <iostream>#include <cstring>using namespace std;void fun(char *src, char *dest);int main(){ char src[11]; while (cin>>src) { char dest[11]; fun(src, dest); if (dest[0] == '0' && strlen(dest) != 1) { cout<<dest+1<<endl; } else { cout<<dest<<endl; } }}void fun(char *src, char *dest){ int j = 0; bool hash[10] = {false}; for (int i = strlen(src) - 1; i >= 0; i--) { if (!hash[src[i]-'0']) { hash[src[i]-'0'] = true; dest[j++] = src[i]; } } dest[j++] = '\0';}14 超长正整数相加
#include <iostream>#include <cstring>#include <string>#include <algorithm>using namespace std;string fun(string n1, string n2);int main(){ string n1, n2; while (cin>>n1 && cin>>n2) { cout<<fun(n1, n2).c_str()<<endl; }}string fun(string n1, string n2){ string n = ""; int num = 0; int i = n1.size() - 1; int j = n2.size() - 1; for (; i >= 0 && j >= 0; i--, j--) { num += n1[i] - '0' + n2[j] - '0'; n += num % 10 + '0'; num /= 10; } while (i >= 0) { num += n1[i--] - '0'; n += num % 10 + '0'; num /= 10; } while (j >= 0) { num += n2[j--] - '0'; n += num % 10 + '0'; num /= 10; } if (num != 0) { n += num + '0'; } reverse(n.begin(), n.end()); return n;}三个循环可以合为一个。
15 图片整理
实际为字符排序。
如果用string str，直接sort(str.begin(), str.end())，需要包含<algorithm>;
#include <iostream>#include <cstring>using namespace std;void sort(char *str);int main(){ char ch[1025]; while (cin>>ch) { sort(ch); cout<<ch<<endl; }}void sort(char *str){ int len = strlen(str); for (int i = 0; i < len - 1; i++) { int j = i; for (int k = i + 1; k < len; k++) { if (str[k] < str[j]) { j = k; } } if (j != i) { char c = str[j]; str[j] = str[i]; str[i] = c; } }}16 在字符串中找出连续最长的数字串
输出最长的数字串及其个数
输入：
abcd123afd1234daf
ab
输出：
1234,4
0
#include <iostream>#include <cstring>using namespace std;unsigned int Continumax(char* pOutputstr, char* pIntputstr);int main(){ char ch[100]; while (cin.getline(ch, 100)) { char output[100] = {0}; int n = Continumax(output, ch); if (strlen(output) == 0) { cout<<n<<endl; } else { cout<<output<<","<<n<<endl; } } return 0;}unsigned int Continumax(char* pOutputstr, char* pInputstr){ int len = strlen(pInputstr); int max = 0; int max_left = 0; int max_right = 0; int left = 0; int right = 0; for (int i = 0; i < len; i++) { if (pInputstr[i] >= '0' && pInputstr[i] <= '9') { right = i; if (right - left + 1 > max) { max = right - left + 1; max_left = left; max_right = right; } } else { left = i + 1; right = i + 1; } } if (max == 0) { return 0; } for (int i = max_left; i <= max_right; i++) { *pOutputstr++ = pInputstr[i]; } *pOutputstr = '\0'; return max;}17 简单密码破解
小写字母对应为手机键盘对应的数字。明码中的大写字母转成原来的小写字母是通过，先变成小写，然后ASCII码再加1。输入为明码，输出为密码。
#include <iostream>#include <cstring>using namespace std;void fun(char *str);int main(){ char ch[100]; while (cin.getline(ch, 100)) { fun(ch); cout<<ch<<endl; } return 0;}void fun(char *str){ int n[26] = {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9}; for (int i = 0; i < strlen(str); i++) { if (str[i] >= 'A' && str[i] <= 'Z') { str[i] = str[i] + 32 + 1; if (str[i] > 'z') { str[i] = 'a'; } } else if (str[i] >= 'a' && str[i] <= 'z') { str[i] = n[str[i] - 'a'] + '0'; } }}18 [中级]单词倒排
将单词倒序排列，中间的空格或其他字符都作为分隔符，输出中的分隔符只能有一个空格，不能有其他符号。
输入：
I am a boy
I 4324 /*- - - /-// fdsf
输出：
boy a am I
fdsf I
#include <iostream>#include <cstring>using namespace std;void fun(char *dest, const char *src);int main(){ char ch[100]; while (cin.getline(ch, 100)) { char dest[100]; fun(dest, ch); cout<<dest<<endl; } return 0;}void fun(char *dest, const char *src){ int len = strlen(src); int i = len - 1; int right = i; int left = i; int count = 0; while (i >= 0) { if ((src[i] >= 'A' && src[i] <= 'Z') || (src[i] >= 'a' && src[i] <= 'z')) { left = i; count++; } else { if (count != 0) { while (left <= right) { *dest++ = src[left++]; } *dest++ = ' '; } left = i - 1; right = i - 1; count = 0; } i--; } //最后的这一个单词最容易忘记 if (count != 0) { while (left <= right) { *dest++ = src[left++]; } } *dest = '\0';}19 输出单向链表中倒数第k个结点
尾结点为倒数第0个。
#include <iostream>using namespace std;struct ListNode{ int m_nKey; ListNode *m_pNext; ListNode(int x) : m_nKey(x), m_pNext(NULL) {}};ListNode* FindKthToTail(ListNode* pListHead, unsigned int k);void print(ListNode *head);int main(){ int n; while (cin>>n) { ListNode *head = new ListNode(-1); ListNode *pre = head; while (n--) { int val; cin>>val; ListNode *cur = new ListNode(val); pre->m_pNext = cur; pre = cur; } int k; cin>>k; ListNode *p = FindKthToTail(head->m_pNext, k); cout<<p->m_nKey<<endl; } return 0;}ListNode* FindKthToTail(ListNode* pListHead, unsigned int k){ ListNode *p1 = pListHead; ListNode *p2 = pListHead; while (k--) { p1 = p1->m_pNext; } while (p1->m_pNext != NULL) { p1 = p1->m_pNext; p2 = p2->m_pNext; } return p2;}20 取近似值
差不多就是四舍五入吧
#include <iostream>using namespace std;int main(){ float n; while (cin>>n) { n >= 0 ? cout<<(int)(n + 0.5)<<endl : cout<<(int)(n - 0.5)<<endl; } return 0;} 0 0
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